A nuclear fuel rod generates energy at a rate of $5 \times 10^{8} \mathrm{Watt} / \mathrm{m}^{3}$. It is in the shape of a cylinder of radius $4.0 \mathrm{~mm}$ and length $0.20 \mathrm{~m}$. A coolant of specific heat $4 \times 10^{3} \mathrm{~J} /(\mathrm{kg}-\mathrm{K})$ flows past it at a rate of $0.2 \mathrm{~kg} / \mathrm{s}$. The temperature rise in this coolant is approximately -

• A

  $2^{\circ} \mathrm{C}$
• Correct answer
  B

  $6^{\circ} \mathrm{C}$
• C

  $12^{\circ} \mathrm{C}$
• D

  $30^{\circ} \mathrm{C}$

Solution

$\frac{\mathrm{dm}}{\mathrm{dt}} \times \mathrm{S} \Delta \mathrm{T}=\frac{\mathrm{d} \theta}{\mathrm{dt}}$ $\frac{\mathrm{d} \theta}{\mathrm{dt}}=5 \times 10^{8} \times$ volume of rod $=5 \times 10^{8} \times \pi \times(4)^{2} \times 10^{-6} \times \frac{0.2}{10}$ $=5 \times 10 \times \pi \times 16 \times 2$ $=1600 \pi$ $0.2 \times 4 \times 10^{3} \Delta \mathrm{T}=1600 \pi$ $8 \times 10^{2} \Delta \mathrm{T}=16 \times 10^{2} \pi$ $\Delta \mathrm{T}=3.14 \times 2$ $\Rightarrow 6.28^{\circ} \mathrm{C}$

Hence, the correct answer is option: 'B'.